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Question
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Solution
\[\text{ Let I }= \int\frac{1}{1 - \cot x}dx\]
\[ = \int\frac{1}{1 - \frac{\cos x}{\sin x}}dx\]
\[ = \int\frac{\sin x}{\sin x - \cos x}dx\]
\[ = \frac{1}{2}\int\frac{2 \sin x}{\sin x - \cos x} dx\]
\[ = \frac{1}{2}\int\left[ \frac{\sin x + \cos x + \sin x - \cos x}{\sin x - \cos x} \right]dx\]
\[ = \frac{1}{2}\int\left( \frac{\sin x + \cos x}{\sin x - \cos x} \right)dx + \frac{1}{2}\int dx\]
\[\text{ Putting sin x }- \cos x = t\]
\[ \Rightarrow \left( \cos x + \sin x \right) dx = dt\]
\[ \therefore I = \frac{1}{2}\int\frac{1}{t}dt + \frac{1}{2}\int dx\]
\[ = \frac{1}{2} \text{ ln }\left| t \right| + \frac{x}{2} + C\]
\[ = \frac{x}{2} + \frac{1}{2} \text{ ln }\left| \sin x - \cos x \right| + C\]
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